If $\vec{a}=\hat{i}+\hat{j}+\hat{k}$ and $\vec{b}=\hat{j}-\hat{k},$ find a vector $\vec{c}$ such that $\vec{a} \times \vec{c}=\vec{b}$ and $\vec{a} \cdot \vec{c}=3.$

(N/A) Let $\vec{c}=x \hat{i}+y \hat{j}+z \hat{k}.$
Given $\vec{a}=\hat{i}+\hat{j}+\hat{k}$ and $\vec{b}=\hat{j}-\hat{k}.$
For $\vec{a} \times \vec{c}=\vec{b},$
$\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ x & y & z \end{vmatrix} = 0\hat{i} + 1\hat{j} - 1\hat{k}.$
Expanding the determinant,we get $\hat{i}(z-y) - \hat{j}(z-x) + \hat{k}(y-x) = 0\hat{i} + 1\hat{j} - 1\hat{k}.$
Comparing coefficients,we have:
$z-y=0 \implies z=y \quad (i)$
$x-z=1 \implies x=z+1 \quad (ii)$
$y-x=-1 \quad (iii)$
Also,$\vec{a} \cdot \vec{c}=3 \implies x+y+z=3 \quad (iv).$
Substituting $x=z+1$ and $y=z$ into $(iv)$:
$(z+1) + z + z = 3 \implies 3z+1=3 \implies 3z=2 \implies z=\frac{2}{3}.$
Thus,$y=\frac{2}{3}$ and $x=\frac{2}{3}+1=\frac{5}{3}.$
Therefore,$\vec{c}=\frac{5}{3}\hat{i}+\frac{2}{3}\hat{j}+\frac{2}{3}\hat{k} = \frac{1}{3}(5\hat{i}+2\hat{j}+2\hat{k}).$